Solve frac{a^{2}+2ab}{a^2+4ab+4b^2}+2b/a-2b-4ab/a^2-4b^2

$$\frac { a ^ { 2 } + 2 a b } { a ^ { 2 } + 4 a b + 4 b ^ { 2 } } + \frac { 2 b } { a - 2 b } - \frac { 4 a b } { a ^ { 2 } - 4 b ^ { 2 } }$$

Evaluate

$$-\frac{2b-a}{a+2b}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{a^{2}+2ab}{a^{2}+4ab+4b^{2}}.

$$\frac{a\left(a+2b\right)}{\left(a+2b\right)^{2}}+\frac{2b}{a-2b}-\frac{4ab}{a^{2}-4b^{2}}$$

Cancel out a+2b in both numerator and denominator.

$$\frac{a}{a+2b}+\frac{2b}{a-2b}-\frac{4ab}{a^{2}-4b^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of a+2b and a-2b is \left(a-2b\right)\left(a+2b\right). Multiply \frac{a}{a+2b} times \frac{a-2b}{a-2b}. Multiply \frac{2b}{a-2b} times \frac{a+2b}{a+2b}.

$$\frac{a\left(a-2b\right)}{\left(a-2b\right)\left(a+2b\right)}+\frac{2b\left(a+2b\right)}{\left(a-2b\right)\left(a+2b\right)}-\frac{4ab}{a^{2}-4b^{2}}$$

Since \frac{a\left(a-2b\right)}{\left(a-2b\right)\left(a+2b\right)} and \frac{2b\left(a+2b\right)}{\left(a-2b\right)\left(a+2b\right)} have the same denominator, add them by adding their numerators.

$$\frac{a\left(a-2b\right)+2b\left(a+2b\right)}{\left(a-2b\right)\left(a+2b\right)}-\frac{4ab}{a^{2}-4b^{2}}$$

Do the multiplications in a\left(a-2b\right)+2b\left(a+2b\right).

$$\frac{a^{2}-2ab+2ba+4b^{2}}{\left(a-2b\right)\left(a+2b\right)}-\frac{4ab}{a^{2}-4b^{2}}$$

Combine like terms in a^{2}-2ab+2ba+4b^{2}.

$$\frac{a^{2}+4b^{2}}{\left(a-2b\right)\left(a+2b\right)}-\frac{4ab}{a^{2}-4b^{2}}$$

Factor a^{2}-4b^{2}.

$$\frac{a^{2}+4b^{2}}{\left(a-2b\right)\left(a+2b\right)}-\frac{4ab}{\left(a-2b\right)\left(a+2b\right)}$$

Since \frac{a^{2}+4b^{2}}{\left(a-2b\right)\left(a+2b\right)} and \frac{4ab}{\left(a-2b\right)\left(a+2b\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{a^{2}+4b^{2}-4ab}{\left(a-2b\right)\left(a+2b\right)}$$

Factor the expressions that are not already factored in \frac{a^{2}+4b^{2}-4ab}{\left(a-2b\right)\left(a+2b\right)}.

$$\frac{\left(a-2b\right)^{2}}{\left(a-2b\right)\left(a+2b\right)}$$

Cancel out a-2b in both numerator and denominator.

$$\frac{a-2b}{a+2b}$$

Expand

$$-\frac{2b-a}{a+2b}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{a^{2}+2ab}{a^{2}+4ab+4b^{2}}.

$$\frac{a\left(a+2b\right)}{\left(a+2b\right)^{2}}+\frac{2b}{a-2b}-\frac{4ab}{a^{2}-4b^{2}}$$

Cancel out a+2b in both numerator and denominator.

$$\frac{a}{a+2b}+\frac{2b}{a-2b}-\frac{4ab}{a^{2}-4b^{2}}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of a+2b and a-2b is \left(a-2b\right)\left(a+2b\right). Multiply \frac{a}{a+2b} times \frac{a-2b}{a-2b}. Multiply \frac{2b}{a-2b} times \frac{a+2b}{a+2b}.

$$\frac{a\left(a-2b\right)}{\left(a-2b\right)\left(a+2b\right)}+\frac{2b\left(a+2b\right)}{\left(a-2b\right)\left(a+2b\right)}-\frac{4ab}{a^{2}-4b^{2}}$$

Since \frac{a\left(a-2b\right)}{\left(a-2b\right)\left(a+2b\right)} and \frac{2b\left(a+2b\right)}{\left(a-2b\right)\left(a+2b\right)} have the same denominator, add them by adding their numerators.

$$\frac{a\left(a-2b\right)+2b\left(a+2b\right)}{\left(a-2b\right)\left(a+2b\right)}-\frac{4ab}{a^{2}-4b^{2}}$$

Do the multiplications in a\left(a-2b\right)+2b\left(a+2b\right).

$$\frac{a^{2}-2ab+2ba+4b^{2}}{\left(a-2b\right)\left(a+2b\right)}-\frac{4ab}{a^{2}-4b^{2}}$$

Combine like terms in a^{2}-2ab+2ba+4b^{2}.

$$\frac{a^{2}+4b^{2}}{\left(a-2b\right)\left(a+2b\right)}-\frac{4ab}{a^{2}-4b^{2}}$$

Factor a^{2}-4b^{2}.

$$\frac{a^{2}+4b^{2}}{\left(a-2b\right)\left(a+2b\right)}-\frac{4ab}{\left(a-2b\right)\left(a+2b\right)}$$

Since \frac{a^{2}+4b^{2}}{\left(a-2b\right)\left(a+2b\right)} and \frac{4ab}{\left(a-2b\right)\left(a+2b\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{a^{2}+4b^{2}-4ab}{\left(a-2b\right)\left(a+2b\right)}$$

Factor the expressions that are not already factored in \frac{a^{2}+4b^{2}-4ab}{\left(a-2b\right)\left(a+2b\right)}.

$$\frac{\left(a-2b\right)^{2}}{\left(a-2b\right)\left(a+2b\right)}$$

Cancel out a-2b in both numerator and denominator.

$$\frac{a-2b}{a+2b}$$